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A category is pseudo-abelian if it is pre-additive and idempotents have images. For my purpose, I only care about additive pseudo-abelian categories, which makes things slightly easier.

I want to show that an idempotent $p : X \to X$ in a pseudo-abelian category splits. Let $(K, k : K \to X)$ be the image of $p$. By the universal property of kernels, there exists a map $r : X \to K$ such that $p = kr$. nLab says that $rk = 1_K$, but I can't prove this.

I have tried to prove this using the uniqueness of the map in the universal property, but this hasn't worked.

Using $p^2 = p$, we get $k(rk) r = kr$. If monics had left inverses and epics right inverse in a pseudo-abelian category (is this true and why?), then we would have $rk = 1_K$.

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    $\begingroup$ Use that $k$ is a mono and $r$ is an epi and $krkr = pp=p= kr=k1_Kr$. i.e $krkr =k1_Kr \Rightarrow rkr = 1_Kr$ (because $k$ is mono) and then $ rkr = 1_Kr \Rightarrow rk=1_K$ (because $r$ is epi). $\endgroup$ – Nex Jun 19 '17 at 9:43
  • $\begingroup$ Yes, of course! This should be an answer. $\endgroup$ – James Jun 19 '17 at 9:47
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    $\begingroup$ By the way, this is nothing special to abelian categories. If an idempotent in any category admits an equalizer with the identity morphism, then that gives a splitting of the idempotent. The only extra thing to show is that the analogue of $r$ is guaranteed to be epi, without assuming it came from an image factorization. $\endgroup$ – Kevin Arlin Jun 20 '17 at 19:24
  • $\begingroup$ Why do you know $r$ is epi? Wikpedia - en.wikipedia.org/wiki/… - says that if C has all equalisers then $r$ is epi. But a pseudo-abelian category might not have all equalisers, and I can't see how to generalise the proof. $\endgroup$ – James Jun 21 '17 at 7:51
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    $\begingroup$ Here is another way that doesn't use that $r$ is epimorphism. I assume that $k$ is the kernel of $1_X-p$. Note that this means that $0=(1_X-p)k=k-pk$ and hence $k=pk$. Now we have that $krk=pk=k$ and so since $k$ is mono $rk=1_K$. $\endgroup$ – Nex Jun 21 '17 at 21:20
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Answer given by Nex in the comments. First a bit of background:

Definition: In a category $\mathscr C$, an idempotent $p : X \to X$ splits if there are morphisms $r : X \to Y$ and $s : Y \to X$ such that $s \circ r = e$ and $r \circ s = 1_Y$. A category $\mathscr C$ is idempotent complete if every non-zero idempotent splits.

Note that the condition $r \circ s = 1_Y$ implies that $r$ is epic and $s$ monic.

Definition: A category $\mathscr C$ is pseudo-abelian if it is additive and every idempotent has an image.

Some authors only require a pseudo-abelian category to be pre-additive.

Every idempotent $p : X \to X$ in an additive category (or a pre-additive category with a terminal object) $\mathscr C$ has a cokernel $(X, 1_X - p)$. The proof of this is straightforward Since $1_X - p$ is also an idempotent if and only if $p$ is an idempotent, an additive category is pseudo-abelian if and only if every idempotent has a kernel.

Proposition: A pseudo-abelian category is idempotent complete.

Proof: Let $p : X \to X$ be an idempotent with image $(K, k : K \to X)$. We know $k$ is the kernel of $1_X - p$. This implies $0 = (1_X-p)k$ and consequently $k = pk$. By the universal property of kernels, there exists a map $r : X \to K$ such that $p = kr$. Thus, $krk = pk = k$ and so $rk = 1_K$ since $k$ is monic.

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